Wave Optics — CBSE Class 12 · 2019 Board Question with Solution

(a) Describe any two characteristic features which distinguish between interference and diffraction phenomena. Derive the expression for the intensity at a point of the interference pattern in Young's double slit experiment.
(b) In the diffraction due to a single slit experiment, the aperture of the slit is

3

mm. If monochromatic light of wavelength

620

nm is incident normally on the slit, calculate the separation between the first order minima and the

3

rd order maxima on one side of the screen. The distance between the slit and the screen is

1.5

Answer: Two features as stated; $I = 4 I_{0} cos^{2} (ϕ /2)$ ; separation $= 0.775$ mm.

Solution

(a) Two features: (i) interference is due to superposition of waves from two coherent wavefronts, diffraction from secondary wavelets of a single wavefront; (ii) interference fringes are equally spaced and equally bright, whereas diffraction gives a dominant central band with weaker, unevenly spaced side bands.

Intensity in YDSE:

let the two waves be

y_{1} = a cos ω t

and

y_{2} = a cos (ω t + ϕ)

. Their resultant is

y = 2 a cos \frac{ϕ}{2} cos (ω t + \frac{ϕ}{2}), A = 2 a cos \frac{ϕ}{2}

Since

I \propto A^{2}

I = 4 I_{0} cos^{2} \frac{ϕ}{2}

(b) For the single slit, the

1^{st}

minimum is at

y_{min} = \frac{λ D}{a}

and the

3^{rd}

secondary maximum at

y_{ma x} = \frac{7}{2} \frac{λ D}{a}

. Their separation is

y = \frac{7}{2} \frac{λ D}{a} - \frac{λ D}{a} = \frac{5}{2} \frac{λ D}{a} = \frac{5}{2} \times \frac{620 \times 1 0 ^{- 9} \times 1.5}{3 \times 1 0 ^{- 3}}

y = 7.75 \times 1 0^{- 4} m = 0.775 mm

y = 0.775 mm

CBSE Class 12 Board · 2019 · Subjective

CBSE Class 12 · 2019 · Wave OpticsLong Answer

Solution

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